A moving boat with a flag on its mast

[brotherbobby]

Homework Statement

A launch travels across a river from point $A$ to point $B$ on the opposite bank along the straight line $AB$ forming an angle $\alpha$ with the bank (see figure below). The wind blows with a velocity $u$ at right angles to the bank. The flag on the mast of the launch forms an angle $\beta$ with the direction of its motion.

(a) Determine the speed of the launch with respect to the bank.

(b) Can the data in this problem be used to find the river current velocity?

Relevant Equations

1. Angles of a triangle add up to $180^{\circ}$.
2. The component of a vector of magnitude $A$ along a direction at an angle $\theta$ to its direction is $A\cos\theta$.

Let me copy and paste the problem as it appeared in the text.




Attempt : First off, the launch can have no velocity of its "own" relative to air current. It is moving entirely due to the air current. Moreover, since no values are given for river, the river current velocity can't be found, which answers (b) above.

I drew a sketch of the problem to the left. The launch moves along AB at an angle $\alpha$ to the horizontal and the flag is along BC at an angle $\beta$ to AB. The wind moves along AC with speed $u$.

We are required to find the speed with which the launch moves along AB, say some $u''=?$

To go there, we realise that the only the velocity of wind perpendicular to the flag of the launch would affect it. Let that speed be some $u'$ shown along AD. Of importance are the angles shown as 1 and 2. We also note that the sum of the two angles $\angle 1+\angle 2 = \angle BAC = \dfrac{\pi}{2}-\alpha$, shown in the diagram.

The approach is to go from $\boldsymbol{u\rightarrow u' \rightarrow u''=?}$

[There's some tedious geometry here, so bear with me. I have marked angle values for help.]

We have $\angle CBA = \pi - \beta\;\text{(shown)}$. In $\small{\triangle ABC, \angle C= \pi-(\pi-\beta)-(\pi/2-\alpha) = \alpha+\beta-\pi/2\;\text{(shown)}}$.
Since $AD\perp BC\Rightarrow \angle D=\pi/2$, we get the value of $\angle 1 = \pi/2-\angle C$, $\angle 1 = \pi/2-(\alpha+\beta-\pi/2)=\pi-(\alpha+\beta)$.

Let me annotate this angle : $\color{blue}{\angle 1 = \pi-(\alpha+\beta)}$.
This would mean the speed $u'=u\cos\angle 1=u\cos[\pi-(\alpha+\beta)]=-u\cos(\alpha+\beta)$.

From the figure again, $\angle 2=(\pi/2-\alpha)-\angle 1=\pi/2-\alpha-[\pi-(\alpha+\beta)]=-\pi/2+\beta$. Let me annonate this angle : $\color{ForestGreen}{\angle 2 = -\pi/2+\beta}$.

Hence the speed along AB : $u''=u'\cos\angle 2=-u\cos(\alpha+\beta)\cos(-\pi/2+\beta)\Rightarrow\boxed{u''=-u\cos(\alpha+\beta)\sin\beta}\quad{\color{red}{\Huge\times}}$

Doubt : My answer doesn't match with the text.

I copy and paste its answer to the right.

Simplifying the text answer, $\boxed{v=-u\dfrac{\cos(\alpha+\beta)}{\sin\beta}}\quad{\color{green}{\huge\checkmark}}$, which means I have only gone wrong in having $\sin\beta$ in the numerator. (I am assuming that, speed being a scalar, the negative sign is of little consequence.)

Request : Where do you think I have gone wrong? Many thanks.

 

[Orodruin]

To go there, we realise that the only the velocity of wind perpendicular to the flag of the launch would affect it.

It is unclear to me what you even mean by this. The flag direction will be the wind direction relative to the ship. This is the wind velocity relative to the bank minus the ship velocity relative to the bank.

 

[brotherbobby]

The flag direction will be the wind direction relative to the ship. This is the wind velocity relative to the bank minus the ship velocity relative to the bank.

Of course I get the second part of your sentence using vector addition.
But it's the first part I don't follow.

The flag has a given direction, at an angle $\beta$ relative to the ship's motion. Are you saying the flag is free to reorient so as to be parallel to the wind relative to the ship?

 

[Orodruin]

Of course I get the second part of your sentence using vector addition.
But it's the first part I don't follow.

The flag has a given direction, at an angle $\beta$ relative to the ship's motion. Are you saying the flag is free to reorient so as to be parallel to the wind relative to the ship?

No. I am saying that this is the direction of the wind relative to the ship.

 

[brotherbobby]

No. I am saying that this is the direction of the wind relative to the ship.

So in the above problem, the flag is directed at an angle $\beta$ relative to the ship's motion. This is also the direction of the wind relative to the ship?

 

[Orodruin]

So in the above problem, the flag is directed at an angle $\beta$ relative to the ship's motion. This is also the direction of the wind relative to the ship?

Yes.

 

[brotherbobby]

Yes.

I think I should re-do the problem along the lines of what you said.

Just an EDIT - I have found the mistake in my solution above in post #1. The speed $u''\ne u'\cos\angle 2,$ as I incorrectly wrote. But rather, from $\triangle ABD$, $u''=\dfrac{u'}{\cos\angle 2}\Rightarrow u''=\dfrac{-u\cos(\alpha+\beta)}{\cos(-\pi/2+\beta)}$, which simplifies to the correct answer :
$\boxed{u''=\dfrac{-u\cos(\alpha+\beta)}{\sin\beta}}$

 

[Orodruin]

Regarding your solution: It is much simpler to just apply the law of sines. It will give you the result directly without further considerations.

 

[brotherbobby]

A few questions remain on this problem, elementary though the problem is.

(1) First off, if wind is blowing vertically "up" with speed $u$, shouldn't the boat also move with the same speed $u$ in the same direction? However, it moves in a different direction with a different speed which I am calling $u''$. Is this because there must be some other velocity on the boat present, say velocity of water current?

(2) Though I could do the problem, I haven't got the role of the flag's direction. A flag or a mast (on sails) we all know what it is. I don't understand the relevance of the flag direction. All I can confirm is that the flag need not be directed in the same direction as the boat moves. What does the direction of the flag have to be with the wind blowing?

Thanks for your interest.

 

[haruspex]

A launch is not wind-powered. The reference to a mast probably means a flagpole.

 

[brotherbobby]

A launch is not wind-powered. The reference to a mast probably means a flagpole.

Yes, it was a mistake in the text to use the term "launch". An admin changed the title of the thread to "boat".

 

[haruspex]

Yes, it was a mistake in the text to use the term "launch". An admin changed the title of the thread to "boat".

Ok, but even a wind-powered boat rarely goes in the direction of the wind, and certainly not at wind speed.
The reason you cannot deduce the current speed is that even if there were no current the boat's speed would be less than the wind speed because of drag from the water.

 

[Orodruin]

1: You seem to be saying a sailing boat has no choice but to move in the wind direction. You only need to watch a sailing contest to know this is inaccurate.

2. You will agree that if you are standing next to a stationary flag pole, the flag will flow in the wind direction, yes? The mast is stationary in the boat frame and the flag will flow in the direction of air flow in that frame.

 

[brotherbobby]

$\small{\texttt{EDIT - I have solved the problem in the more matured way using relative vectors.}}$
$\small{\texttt{I state the problem statement first and then go about solving it.}}$

Problem statement :





Solution : (b) is NO, obviously, because no data exists. However, that the boat is moving in a direction different from wind velocity vector $\vec u$ shows that water current is present. In that sense, the correct question to ask in (a) would be : Determine the speed of the boat with respect to the waster current.

I draw a sketch of the problem. We need to find the velocity of the boat relative to water, shown as $\color{red}{u''=?}$ and drawn with a red line ($\color{red}{\rule{10mm}{0.8pt}}$). This speed is at an angle of $\alpha$ to the horizontal, or at an angle $\dfrac{\pi}{2}-\alpha\; {\text{(shown)}}$ to the vertical. The wind moves with a velocity $\color{brown}u$ vertically up, shown with a brown line ($\color{brown}{\rule{10mm}{0.8pt}}$). The flag is shown with a green line ($\color{YellowGreen}{\rule{10mm}{0.8pt}}$). It is directed at an angle of $\beta$ to the velocity vector of the boat $u''$. Hence the angle inside the triangle is $\pi-\beta\;\text{(shown)}$.

$\small{\boxed{\texttt{Main concept}} : \texttt{Direction of flag = Direction of wind relative to motion}}$

Let the velocity of wind relative to motion be $\color{YellowGreen}{u'}$. It is directed along the flag.

Let us focus on the triangle formed by the three vectors $u, u', u''$. Two of the angles are known and shown. The third can be found by the angle sum property and is $\alpha+\beta-\dfrac{\pi}{2}$. Let us focus on the sides $u''$ and $u$ and their opposite angles. Using the Sine Rule :
$\small{\dfrac{u''}{\sin(\alpha+\beta-\pi/2)}=\dfrac{u}{\sin(\pi-\beta)}}\Rightarrow\boxed{u''=-u\dfrac{\cos(\alpha+\beta)}{\sin\beta}}$

 

[brotherbobby]

1: You seem to be saying a sailing boat has no choice but to move in the wind direction. You only need to watch a sailing contest to know this is inaccurate.

I understand, but that makes the whole problem a tricky one. Let's stick to the simpler situation - the boat moves with the same velocity as the wind which means the water offers no resistance.

2. You will agree that if you are standing next to a stationary flag pole, the flag will flow in the wind direction, yes? The mast is stationary in the boat frame and the flag will flow in the direction of air flow in that frame.

Yes. This is the very concept on which the weathercock or wind vane on the top of some houses is. Being a new idea for me, I have to get used to it. One way out is to argue what happens if there was no wind. If you held a flag and walked with velocity $\vec v$, the flag would point in the direction $-\vec v$, which is the same direction in which wind would flow relative to you. Now, if there was wind blowing with some velocity $\vec u$, the flag would point in the direction of the resultant $-\vec v+\vec u$.

 

[Orodruin]

You cannot go around making random assumptions about a posed problem. The problem is what it is and the assumption of the motion direction does not make it harder or easier in any way or form.